\begin{problem*}{10.21}
Find the net torque $\tau$ on the wheel in Fig. P10.21 about the axle
through $O$, taking $a = 10.0\U{cm}$ and $b = 25.0\U{cm}$.
\end{problem*} % problem 10.21

\begin{solution}
Torque is defined as $\vect{\tau} = \vect{r} \times \vect{F}$, so we
have (defining the counter clockwise direction to be positive)
\begin{equation}
 \sum \tau = - b \cdot 10.0\U{N}  + a \cdot 12.0\U{N} - b \cdot 9.00\U{N}
\end{equation}
Where the $-$ sign on the first and third terms denote torques in the
$-$ direction.  There are no $\sin$ terms, because all three forces
are in the tangential direction.  The $12.0\U{N}$ force is slightly
suspicious, since they tell you it makes an angle of $30\dg$ with the
horizontal, but if you look closely, you'll see that it isn't actually
applied to the top of the circle, and it {\it is} tangential to it's
application radius.

Plugging in for $a$ and $b$ we have
\begin{equation}
 \sum \tau = -0.250\U{m} \cdot 10.0\U{N} + 0.100\U{m} \cdot 12.0\U{N} - 0.250\U{m} \cdot 9.00\U{N}
           = \ans{-3.55\U{J}}
\end{equation}
\end{solution}
